Question

假设随机变量X和Y相互独立,服从标准正态分布,求随机变量Z=X/Y的概率密度。求详细过程

Answer 1

联合密度函数f(x,y)=f(x)*f(y)=(1/2π)e^[-(x^2+y^2)/2]
画图可知（X为纵坐标，Y为横坐标）
是的Z<z的区域是Y=0和X=Yz所夹的区域
所以FZ(z)=P{Z<=z}
=∫∫(D(z))f(x,y)dxdy
做极坐标变换
y=rcosθ，x=rsinθ
则0<r<+∞，(π/2)<θ<(π/2)+arctanz或(3π/2)<θ<3π/2
+arctanz
FZ(z)=(∫(π/2,π/2+arctanz)+∫(3π/2,3π/2
+arctanz))dθ
∫(0,+∞)
r
*
(1/2π)
*
e^(-r^2/2)dr
=(π+2arctanz)
*
∫(0,+∞)
(1/2π)e^(-r^2/2)d(r^2/2)
=(1/2)
+
(1/π)arctanz
所以fZ(z)
=
(1/π)/(1+z^2)

Answer 2

可能你想写：求
z=(x^2＋y^2)^0.5的密度函数.
f(z)=p(z<=z)=p{(x^2+y^2)^0.5<=z}
当z<0时，f(z)=0
当z>=0时，f(z)=p{x^2+y^2<=z^2}
f(z)=p{x^2+y^2<=z^2}=(2πσ^2)^(-1)∫∫e^[-(x^2+y^2)/(2σ^2)]dxdy,
积分区域是x^2+y^2<=z^2
积分得概率分布：f(z)=1-e^[-z^2/(2σ^2)]，z>0
求导得概率密度：f(z)=(z/σ^2)*e^[-z^2/(2σ^2)],z>=0,f(z)=0,z<0